In Freundlich adsorption isotherm, the value of \(\frac{1}{n}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → between 0 and 1 in all cases
The Freundlich adsorption isotherm gives the extent of adsorption x/m of a gas on a solid at a given temperature:
\( \displaystyle\frac{x}{m} = k\,p^{1/n}\ \ (n > 1), \qquad \log\displaystyle\frac{x}{m} = \log k + \displaystyle\frac{1}{n}\log p \)
The factor 1/n can have values between 0 and 1 (the slope of the log plot):
1/n = 0: x/m = constant; adsorption is independent of pressure (at high pressure, surface saturated).
1/n = 1: x/m = kp; adsorption varies directly with pressure (at low pressure).
The isotherm applies to physisorption and chemisorption alike; 1/n is not fixed at 1 for either.
Why the other options are incorrect
Option (1) 1 in case of chemisorption — Incorrect. 1/n is not tied to the type of adsorption; it depends on pressure and the system.
Option (2) 1 in case of physisorption — Incorrect. 1/n = 1 only describes the low-pressure limit, not physisorption in general.
Option (4) between 2 and 4 in all cases — Incorrect. Since n > 1, 1/n is always less than 1; a value of 2 to 4 is impossible.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Adsorption Isotherms – “… Freundlich isotherm explains the behaviour of adsorption in an approximate manner. The factor 1/n can have values between 0 and 1 (probable range 0.1 to 0.5). Thus, equation (5.2) holds good over a limited range of pressure.”